I have to find the $Y$ of first half, where, for $\text{id} / 2$, $Y = 100$.
$Y$ has to follow up, so that in the end totals of input match the totals of output.
For first half, each rows output has to be higher than input.

How do I do this, and, how would this break into steps?

P.S. The tag might be wrong, I have no idea about the English mathematical terminology, not my native language.

update

To narrow it down a bit (previous table with limit of 10, filled in, explained):

1 Answer
1

I don't really understand your question, so with your help I'll try to clarify it, for myself really.

I'd write, for clarity, the id as $i$, the input as $x_{i}$, output as $y_{i}$ and percentages as $p_{i}$. Then $y_{i}=p_{i} * x_{i}$ and $\sum_{i=1}^n x_{i} = \sum_{i=1}^n y_{i}$. This is what I've managed to gather so far.

Now, what did you mean by these 2 statements:

Ending half of the table (> id / 2 < n), Y = 50.

I have to find the Y of first half, where, for id / 2, Y = 100.

Is it that for all $i$, when $\frac{i}{n} &lt \frac{1}{2}$, then $p_{i}=100$ and when $\frac{i}{n} > \frac{1}{2}$, then $p_{i}=50$ ? Also, I think that this condition is also necessary: $\sum_{i=1}^n p_{i}=1$.

@Tom: I made the necessary changes, after having noticed $y_{1}>y_{2}>y_{3}>y_{4}>100$.
–
leqsApr 26 '12 at 16:51

Does not make sense if $y_{2}$ < $x_{2}$, but I see the problem. I have specified the percentages as Y, where you are using $y$ for output. With that in mind, it's not $y_{1} > y_{2} > y_{3} > y_{n} > 100$, but $p_{1} > p_{2} > p_{n} > 100$ (or in decimals, $> 1.00$), that results always in $y > x$, but not always $y_{n} > y_{n-1}$ (the first doesn't get the highest output, but the highest percentage). P.S. Excuse my bad explaining, it's just that numbers are killing me, even theoretically, therefore I end up with complicated explanations.
–
joltmodeApr 27 '12 at 7:09